PHP - SQL making a menu with submenu's

Question

Hy,

The title says it al, i want to make a menu with submenu's.

So something like this:

parent1
1. sub2
  1. sub 3
parent2

I've got this already working, but the code doesn't look nice.

So how can i improve my code, or does you know it beter, post it then.

This is the code:

<?php

$le_content .= '<div id="main_menu">';


$sQuery_get_all_parents = " SELECT
                            menu.id,
                            menu.`name`,
                            menu.url,
                            menu.parent_id
                            FROM `menu`
                            WHERE
                            menu.actif = 1 AND
                            menu.parent_id = 0
                            ORDER BY
                            menu.volgorde ASC";
$rResultaat_get_all_parents = mysql_query($sQuery_get_all_parents) or die(mysqlError($sQuery_get_all_parents));

$menu_parents = array();
$menu_child_1 = array();
$counter = 0;
$counter_child_1 = 0;
$counter_child_2 = 0;

while($row = mysql_fetch_assoc($rResultaat_get_all_parents)){
    $menu_parents[$counter]['id'] = $row['id'];
    $menu_parents[$counter]['name'] = $row['name'];
    $menu_parents[$counter]['url'] = $row['url'];
    $menu_parents[$counter]['parent_id'] = $row['parent_id'];
    $counter++;
}

$le_content .= '<ul id="main_menu_ul">';
foreach ($menu_parents as $menu_entry => $value) {
    $le_content .= '<li><a href="'.ROOT_HREF.$value['url'].'" >'.$value['name'].'</a>';
    $sQuery_get_first_children = "  SELECT
                                    menu.id,
                                    menu.`name`,
                                    menu.url,
                                    menu.parent_id
                                    FROM `menu`
                                    WHERE
                                    menu.actif = 1 AND
                                    menu.parent_id = ".$value['id']."
                                    ORDER BY
                                    menu.volgorde ASC";
    $rResultaat_get_first_children = mysql_query($sQuery_get_first_children) or die(mysqlError($sQuery_get_first_children));
    $row_count = mysql_num_rows($rResultaat_get_first_children);

    if($row_count != false){
        $le_content .= '<ol>';
        while($row = mysql_fetch_assoc($rResultaat_get_first_children)){
            $menu_child_1[$counter_child_1]['id'] = $row['id'];
            $menu_child_1[$counter_child_1]['name'] = $row['name'];
            $menu_child_1[$counter_child_1]['url'] = $row['url'];
            $menu_child_1[$counter_child_1]['parent_id'] = $row['parent_id'];
            $counter_child_1++;
        }

        foreach ($menu_child_1 as $menu_entry_child_1 => $value_child_1) {
            $le_content .= '<li><a href="'.ROOT_HREF.$value_child_1['url'].'" >'.$value_child_1['name'].'</a>';
            $sQuery_get_second_children = " SELECT
                                            menu.id,
                                            menu.`name`,
                                            menu.url,
                                            menu.parent_id
                                            FROM `menu`
                                            WHERE
                                            menu.actif = 1 AND
                                            menu.parent_id = ".$value_child_1['id']."
                                            ORDER BY
                                            menu.volgorde ASC";
            $rResultaat_get_second_children = mysql_query($sQuery_get_second_children) or die(mysqlError($sQuery_get_second_children));
            $row_count = mysql_num_rows($rResultaat_get_second_children);
            if($row_count != false){
                $le_content .= '<ol>';
                while($row = mysql_fetch_assoc($rResultaat_get_second_children)){
                    $menu_child_2[$counter_child_2]['id'] = $row['id'];
                    $menu_child_2[$counter_child_2]['name'] = $row['name'];
                    $menu_child_2[$counter_child_2]['url'] = $row['url'];
                    $menu_child_2[$counter_child_2]['parent_id'] = $row['parent_id'];
                    $counter_child_2++;
                }
                foreach ($menu_child_2 as $menu_entry_child_2 => $value_child_2) {
                    $le_content .= '<li><a href="'.ROOT_HREF.$value_child_2['url'].'" >'.$value_child_2['name'].'</a></li>';
                }
                $le_content .= '</ol>';
            }

            $le_content .= '</li>';
        }
        $le_content .= '</ol>';
    }
    $le_content .= '</li>';
}

$le_content .= '</ul></div>';

?>

in the end, i echo $le_content.....

Please consider switching to MySQLi or PDO. The mysql_* functions are old, soon to be deprecated, and don't provide solid counter-measures against SQL injection attacks. — Polynomial
– Polynomial, Commented Jul 2, 2012 at 14:26

Peter Kiss · Accepted Answer · 2012-07-02 14:21:04Z

4

Query all menu item
Build a hierarchical array of them (create a recursive function to do it) every menuitem should have a property for it's children collection
Create another recursive function to write the output

answered Jul 2, 2012 at 14:21

Peter Kiss

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Comments

Mathlight · Accepted Answer · 2012-07-02 20:46:09Z

0

Alright, i've searched around, and came with an solution... Making the query better. i've followed this tutorial and that works verry well:

sql and categories

Thanks for all the help ;)

answered Jul 2, 2012 at 20:46

Mathlight

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